Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Cumulative Review - Page 473: 47a

Answer

$\dfrac{1}{8}$

Work Step by Step

Using $a^{-x}=\dfrac{1}{a^x}$ and $a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$, the given expression, $ 16^{-3/4} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{16^{3/4}} \\\\= \dfrac{1}{(\sqrt[4]{16})^{3}} \\\\= \dfrac{1}{\left( \sqrt[4]{(2)^4} \right)^{3}} \\\\= \dfrac{1}{2^{3}} \\\\= \dfrac{1}{8} .\end{array}
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